Second derivative as a multilinear map

Question

Well, let's assume that $T=(f\circ g)$. How do I compute $D^2 T(y)(v,w)$?

I tried $D^2 T(y)(v,w)=(D_wD_vT)(y)$, i.e. do first the directional derivatives and then compute them at $y$.

$D_vT=((Df)\circ g) \circ (Dg )\ (v)$, then I try to apply again the chain rule to $D_wD_vT$.

My best (incorrect) effort gives $((D((Df)\circ g))\circ Dg \ (v))\circ D^2g(v,w)=((D^2f)\circ g)\circ Dg )\circ Dg \ (v))\circ D^2g(v,w)$ which is a big mess and this doesn't look like what I'm supposed to get, which is $$D^2f(x)(Dg(y)v,Dg(y)w)+Df(x)D^2g(y)(v,w)$$

How would one proceed? any help would be appreciated.

P.S: If you're downvoting this question, at least tell me what's wrong it with so that I can improve it. Thanks.

Answer 1

Notice that for a continuous bilinear map $\mathcal B : X\times Y \to Z$ on normed spaces $X,Y,Z$ we have $$ D\mathcal B(x, y)[h, k] = \mathcal B(h, y) + \mathcal B(x, k). $$ This can be directly shown using the definition of frechet differentiability.

Now, the space of continuous linear maps is again a normed space and the composition / multiplication of compatible continuous linear maps is bilinear and continuous. For $A(y) = Df(g(y))$ and $B(y) = Dg(y)$ we have \begin{align*} D^2 T(y)[w] &= D( A(y) \, B(y) )[w] \\ &= DA(y)[w] \, B(y) + A(y) \, DB(y)[w] \\ &= D^2 f(g(y))[Dg(y)[w],\, Dg(y) \,\cdot\,] + Df(g(y)) \, D^2g(y)[w,\,\cdot\,]. \end{align*}

Notice that the frechet differential of order $2$ is a symmetric bilinear map. Thus, the order of the linear arguments of $D^2 f$ and $D^2 g$ is arbitrary.

Additional Notes:

Say, we have $f: Z \to E$, $g: Y \to Z$ and $T = f\circ g: Y \to E$.

1. $D^2 f$ is recursively defined as $D(Df)$. Here, $Df$ maps into $L(Z; E)$, space of continuous linear map from $Z$ into $E$. So, $D^2 f$ maps into $L(Z; L(Z;E))$ (isomorph to continuous bilinear maps $L(Z, Z; E)$). Now, with mild assumptions on $E$, $D^2 f(z)$ is in fact symmetric, that is $$ (D^2 f(z)v)w = (D^2 f(z)w)v =: D^2 f(z)[v, w] $$ holds for every $v,w\in Z$.

2. Now, consider $A = Df\circ g: Y\to L(Z;E)$ and $B = Dg: Y\to L(Y;Z)$. By chain rule we have $$ DA(y)v = D^2 f(g(y))Dg(y)v \in L(Y;E) $$ and $$ ((DA(y)v) \, B(y)) \,w = ((D^2 f(g(y))Dg(y)v) \, B(y)) \, w = D^2 f(g(y))[Dg(y) v, B(y) w]. $$ That doesn't rely on the symmetry of $(v,w)\mapsto (DA(y)v)w$. One just need to track the order of the arguments more carefully.